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Math Help - Inplicit Differentiation Problem

  1. #1
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    Thumbs up Inplicit Differentiation Problem

    The problem is to prove that \frac{dy}{dx}=\frac{-1}{(x+1)^2} when  x\sqrt{y+1}+y\sqrt{x+1}=0. I tried substituting x=\tan^2\theta and y=\tan^2\phi but then I got stuck. I tried many other things but this was the only notable one.


    thanks.
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  2. #2
    Member sbhatnagar's Avatar
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    Re: Inplicit Differentiation Problem

    Do you know about the chain rule?
    For example: Solve \frac{d}{dx}(\sqrt{x+1})
    Let y=\sqrt{x+1} and u=x+1

    Then: \frac{dy}{dx}=\frac{du}{dx}\frac{dy}{du}=(1)(\frac  {1}{2\sqrt{u}})=\frac{1}{2\sqrt{x+1}}
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  3. #3
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    Re: Inplicit Differentiation Problem

    Yes, I know the basic formulae and algebra of derivatives.
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  4. #4
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    Re: Inplicit Differentiation Problem

    Just in case a picture helps...



    ... where (key in spoiler) ...

    Spoiler:


    ... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to the main variable (in this case x), and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).

    But this is wrapped inside the legs-uncrossed version of...



    ... the product rule, where, again, straight continuous lines are differentiating downwards with respect to x.


    Larger


    Solve the bottom row (for dy/dx).

    Edit: ... which is easier said than done. Wolfram doesn't have any better ideas: differentiate x sqrt(y+1) + y sqrt(x+1) = 0 with respect to x - Wolfram|Alpha.

    Ok, I'm beginning to see the tan^2 idea...

    ... nah! Was it suggested? What is the context?
    Last edited by tom@ballooncalculus; November 19th 2011 at 06:05 AM.
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  5. #5
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    Re: Inplicit Differentiation Problem

    I took the original equation, moved one term to the RHS, squared both sides of the resulting equation and arrived at this quadratic equation ...

    (x+1)y^2 - x^2y - x^2 = 0

    using the quadratic formula ...

    y = \frac{x^2 \pm \sqrt{x^4 - 4(x+1)(-x^2)}}{2(x+1)}

    which simplifies to either y = x (which, by inspection, does not work in the original equation) or y = \frac{-x}{x+1}

    the derivative of the second equation is y' = \frac{-1}{(x+1)^2}

    might be some other way, but I don't see it ...
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  6. #6
    MHF Contributor FernandoRevilla's Avatar
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    Re: Inplicit Differentiation Problem

    Quote Originally Posted by skeeter View Post
    might be some other way, but I don't see it ...
    You have chosen the best way in this case.
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  7. #7
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    Re: Inplicit Differentiation Problem

    I agree with FernandoRevilla. this is the best way. thanks again.
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